Recursive function to check ascendingly sorted array

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the program is supposed to be a recursive function to check ascendingly sorted array
here's my trial but the program doesn't run well it's supposed to print 1 ,instead, it prints -1!

What I have tried:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int iss(int list[],int size)
{
    if(*list > *(list+1)) return -1;
    if(list==list+size-1) return 1;
    return iss(list+1,size);
}

int main()
{ int arr[5]={1,1,5,6,7};
printf("%d",iss(arr,5));
return 0;
}

Posted 4-Nov-17 9:07am

Member 13476370

Updated 4-Nov-17 11:39am

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PIEBALDconsult 4-Nov-17 15:53pm

For that task; iterate, don't recurse.
And if you know it's sorted, then try a binary search.

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CPallini · Accepted Answer · 2017-11-04T11:39:00

Solution 1

The following stop condition

Quote:
f(list==list+size-1) return 1;

is flawed: if, at first iss call, (size > 1) then it can be never satisfied.
Try instead

C

int iss(int list[],int size)
{
    if ( size == 0 ) return 1;
    if(*list > *(list+1)) return -1;
    return iss(list+1,size-1);
}

Posted 4-Nov-17 11:39am

CPallini

Comments

Member 13476370 5-Nov-17 0:02am

I understand your solution and it's surely better than mine ... but I still don't know why can't if(list==list+size-1) be satisfied? ... every time I change the pointer arithmetically to the next element then The last pointer points at last element of the array ... why isn't that supposed to work?

CPallini 5-Nov-17 6:07am

Think about:
list == list + size - 1
on the left and right side of the expression the value of list is always the same, hence the whole expression is equivalent to
0 == size - 1
since size doesn't change, if at first call (size > 1) then the result of the expression can never be true.